Physics · JEE | Current Electricity
हिन्दीThe rate of electrical energy dissipated as heat in a resistor of resistance \( R \) carrying a current \( \varkappa \) (or \( I \)) is expressed as:
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Correct answer · Option D
\( P = I^2 R = \frac{V^2}{R} \)
Explanation
Using \( P = VI \) and Ohm's law \( V = IR \), power loss is \( P = I^2 R = \frac{V^2}{R} \).